Prove that the probability of occurrence of all of a given set of events A(1) through A(n) is equal to the sum of the individual probabilities minus the sum of the probabilities of all pairs of events, A(i) OR A(j) plus the sum of all triples of events, A(i) OR A(j) OR A(k), ..., plus (-1)^(n-1) times the n-tuple A(i) OR ... OR A(n).

Prove for the specific cases of n = 3 and n = 10, and the general case.

Well I can explain what I was thinking about this... It's far from a proof...

With a 3 set:

Start with nothing:

Add in with at least 1 in common:
A(x)+A(y)+A(z)

Subtract with at least 2 in common:
-(A(x*y)+A(x*z)+A(y*z))

Add in with at least 3 in common:
+(A(x*y*z))



Add in with an odd number in common, subtracting in with an even number in common is an algorithm that gives the probability of A(x) OR A(y) OR A(z) OR ...

Does this look ok, and can anyone branch off of it?

Posted by Gamer on 2003-03-15 03:05:06